Prokaryotes replicate by copying their DNA and splitting their cell into two. The process resembles mitosis but has far fewer steps. Due to this prokaryotes replicate at a much faster rate.

Although prokaryotes generally replicate faster than eukaryotic cells, the time taken to create a new generation varies from prokaryote to prokaryote. Scientists measure the replication rate of different prokaryotes by measuring their generation time which is also known as doubling time.

Some bacteria such as *E. Coli* will double in around 20 minutes while others will take 5-10 hours to produce a new generation. The MCAT will occasionally ask us to use this information to calculate the number of cells after a period of time. These types of problems are identical to half-life problems seen in the chemical and physical foundations section of the exam with one exception. They go backwards in a half-life chain.

Since doubling time and half-life share the same problem setup we can apply what we have already learned about nuclear decay to doubling time. Previously our goal was to find the number of half-lives then work towards what the question asks about. We will focus on the same general goal here, except we will be looking for the number of generations rather than half-lives.

Let’s look at an example problem to see how this works.

Scientists experimentally determined the generation time of *S. Pyogenes* in culture by plating 10 colony forming units (CFUs) onto an agar medium. They then counted the number of CFUs every hour and plotted the data (Figure. 1).

Given the data how many CFUs will be present after 47 hours if a plate started with 2 CFUs of *S. Pyogenes*?

As with radioactive decay questions our goal is to get to the total number of generations in the 47 hour period. To find this information we need the doubling time or how long it takes for the bacteria to double. In this question the doubling time is found from the graph. Since we started with 10 CFUs we will want to find the amount of time it took to get to 20 CFUs since that is double what we started with.

From the graph, we can see that it took about 12 hours for the bacteria to double. Now that we have this information we can find the number of times the bacteria doubled in the 47 hour period. Since 47 isn’t easily divisible by 12 I am going to round 47 to 48.

[latexpage]

\[ \frac{Generation}{12\; hours}\times \frac{48\;hours}{1} \to 4 \; Generations\]

So in a 47 hour period the bacteria will double 4 times. To solve for the total number of CFUs present draw out 4 arrows with blanks in between and a 2 at the very beggining as shown below.

2→ → → →

Now double each number at the beginning of an arrow until you have a final number that corresponds to the total number of CFUs after the 47 hour period.

2→4→8→16→32

So in total, there will be roughly 32 CFUs after the 47-hour time frame. The arrow method works well for 5 or fewer generations but what if you need to determine more than that?

Say the question asked for the number of CFUs present after 12 generations instead. This time it is impractical to draw out 12 arrows instead we need to recognize that each arrow represents multiplication by 2. Therefore we can multiply our original number by 2^{x} where x is the number of generations.

So if we wanted to know how many CFUs would be present after 12 generations if we started with 4 CFUs we would multiple 4 times 2^{12}. Uhhh Chris I don’t know what 2^{12} is! Me neither, but we can figure it out.

[latexpage]

By the exponent rules:

\[ 2^{12} = 2^3\times 2^3\times 2^3\times 2^3\]

Since

\[2^3 = 8\]

\[2^{12}=8\times8\times8\times8 \to 64\times 64 \to 4096\]

Now we can figure out the amount of CFUs by multiplying the above number by the starting cell amount. To make the math easier we will round to 4100.

\[CFUs=4\times 4100 \to 16400\]

So our 4 CFUs will be 16,400 CFUs after 12 generations.

From the doubling time example above we have seen that bacterial division and growth is exponential. However bacteria out in the real world don’t just double forever. Furthermore, bacteria don’t just start growing the minute you put them in the media they have to adapt first.

This leads to a characteristic growth pattern for bacteria that begins with bacteria adjusting to their new environment. This phase is called the lag phase and represents bacteria generating the proper enzymes and figuring out how to acquire food from the environment.

Once the bacteria have adapted they begin doubling and grow exponentially in the exponential phase. At some point in time the bacteria use up all of the available resources and stagnate in the stationary phase.

After a period of time, the bacteria begin to die rapidly as the bacteria are no longer able to survive on the small reserves they had left. This period of rapid death is called the death phase. Bacterial populations can rebound from this phase if additional nutrients are added to their environment however in cell cultures the bacteria typically die out as the food resources are exhausted.

During binary fission prokaryotes faithfully replicate their genomes to the best of their ability. Mutations creep in from time to time, but these mutations wouldn’t account for a bacteria’s ability to suddenly acquire antibiotic resistance. Instead, bacteria have small circular bits of extrachromosomal DNA called plasmids that allow them to acquire new traits rapidly. Plasmids can be acquired from a variety of different sources and in special circumstances even integrated into the genome.

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